Time and Work

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A work is done by 30 workers not all of them have the same capacity to work. Every day exactly 2 workers, do the work with no pair of workers working together twice. Even after all possible pairs have worked once, all the workers together works for two more days to finish the work. Find the number of days in which all the workers together will finish the whole work?









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Option(A) is correct

30 workers work in pairs, with no same pair of workers working together.

Each worker will be working with other 29 which means each worker will work for 29 days in pair.

Let the time taken by each worker be $W_1$, $W_2$, $W_3$..........$W_{30}$ respectively

According to Question

{work done when the workers work in pairs}+{work done when all the workers work together for two days} = 1


\(\Rightarrow 31\left[\dfrac{1}{W_1}+\dfrac{1}{W_2}+\dfrac{1}{W_3}......\dfrac{1}{W_{30}}\right]=1\)

\(\Rightarrow \left[\dfrac{1}{W_1}+\dfrac{1}{W_2}+\dfrac{1}{W_3}......\dfrac{1}{W_{30}}\right]=\dfrac{1}{31}\)

If all the workers work together they will finish the whole work in 31 days.

(3) Comment(s)


30 workers all possible pairs 30c2 =435 so that days,and 30 workers worked for 2 days so total work for 1 person 2*435+30*2=930 is the total work for 30 workers it will take 930/30=31


If each worker is working with every other once shouldn't the number of days be $\dfrac{n(n-1)}{2}+2$


I don't undeerstand the logic behind this? Would you please elaborate a litle more?